9–61. Evaluate and simplify y'.

y=sin √cos² x+1

Verified step by step guidance

First, identify the given function: \( y = \sin(\sqrt{\cos^2 x + 1}) \). We need to find the derivative \( y' \).

Recognize that this is a composition of functions, so we will use the chain rule. The outer function is \( \sin(u) \) where \( u = \sqrt{\cos^2 x + 1} \).

Differentiate the outer function \( \sin(u) \) with respect to \( u \), which gives \( \cos(u) \).

Next, differentiate the inner function \( u = \sqrt{\cos^2 x + 1} \) with respect to \( x \). This requires using the chain rule again: first differentiate \( \sqrt{v} \) where \( v = \cos^2 x + 1 \), giving \( \frac{1}{2\sqrt{v}} \).

Finally, differentiate \( v = \cos^2 x + 1 \) with respect to \( x \). This involves using the chain rule on \( \cos^2 x \), which gives \( -2\cos(x)\sin(x) \). Combine all these derivatives using the chain rule to find \( y' \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Chain Rule

The Chain Rule is a fundamental differentiation technique used to differentiate composite functions. It states that if a function y is composed of another function u, then the derivative of y with respect to x can be found by multiplying the derivative of y with respect to u by the derivative of u with respect to x. This is essential for evaluating derivatives of functions like y = sin(√(cos²(x) + 1), where multiple layers of functions are involved.

Trigonometric Derivatives

Trigonometric derivatives refer to the derivatives of trigonometric functions such as sine, cosine, and tangent. For instance, the derivative of sin(u) is cos(u) multiplied by the derivative of u. Understanding these derivatives is crucial for simplifying expressions involving trigonometric functions, especially when applying the Chain Rule in the given problem.

Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations where y is not explicitly solved for x. It allows us to find the derivative of y with respect to x by treating y as a function of x, even if y is defined implicitly. This concept is useful when dealing with complex functions, such as the one in the question, where direct differentiation may not be straightforward.

9–61. Evaluate and simplify y'.y=sin √cos² x+1